+0

0
94
4

Find the value of $$x = 1 + \cfrac{1}{2 + \cfrac{1}{2 + \cfrac{1}{2 + \cfrac{1}{2 + \ddots}}}}$$

Hint(s):What's the relationship between $$x$$ and $$x+1$$?

Dec 21, 2018

#1
0

Here is a computer summation of your continued fraction:
=1E100; listforeach(b,reverse(1,2,2,2.........etc ), c=b + 1/c);printc
c=1.4142135623730950488016887242097

The result is sqrt(2)

Dec 21, 2018
edited by Guest  Dec 21, 2018
edited by Guest  Dec 21, 2018
#2
+99327
+2

$$x = 1 + \cfrac{1}{2 + \cfrac{1}{2 + \cfrac{1}{2 + \cfrac{1}{2 + \ddots}}}}\\ x = -1+2 + \cfrac{1}{2 + \cfrac{1}{2 + \cfrac{1}{2 + \cfrac{1}{2 + \ddots}}}}\\ x +1= 2 + \cfrac{1}{2 + \cfrac{1}{2 + \cfrac{1}{2 + \cfrac{1}{2 + \ddots}}}}\\ x +1= 2 + \cfrac{1}{x+1}\\$$

$$x +1= 2 + \cfrac{1}{x+1}\\ x-1= \cfrac{1}{x+1}\\ x^2-1=1\\ x^2=2\\ x=\pm\sqrt2$$

I do not know why the answer cannot be -sqrt2  but if memory serves me correctly the only correct answer is +sqrt2

I'll continure thinking about that one.

Dec 21, 2018
#3
+98130
+2

It cannot be negative because the right side of the original equation is positive......thus...x is positive

CPhill  Dec 21, 2018
#4
+99327
0

Yes, I guess so.  Thanks Chris.

Melody  Dec 22, 2018